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Number of ways to line up $5$ males and $5$ females such that no two males are standing next to eachother

I was never good at counting arguments like these! If somebody could explain to my where my argument fails and how to do this correctly I'd really appreciate it.

Since no two males can be standing next to each other, and there are the same number of females as males, the line must alternate between male and female. Therefore the line is going to look either like:

$MFMFMFMFMF$

or

$FMFMFMFMFM$

Now, the number of ways to arrange the 5 males is $5!$, and the number of ways to arrange the number of females is $5!$. Since we could start either with a male or female, the total number of ways to line these people up is:

$(5!)(5!)(2)$

However, I'm pretty sure the correct answer is $(6!)(5!)$

Help appreciated! Thanks!

1 Answers1

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You counted cases in which no two males and no two females are standing next to each other. However, it is possible for two females to stand next to each other. Line up the five females, which can be done in $5!$ ways. This creates six spaces in which to place the five males. $$\square F_1 \square F_2 \square F_3 \square F_4 \square F_5 \square$$ To ensure that no two males stand next to each other, choose five of these six spaces in which to place the males. The males can then be arranged in the five selected spaces in $5!$ ways. Hence, there are $$5!\binom{6}{5}5! = 5! \cdot \frac{6!}{5!1!} \cdot 5! = 5! \cdot \frac{6!}{1!} = 5! \cdot \frac{6!}{1} = 5!6!$$ arrangements in which no two males are adjacent.

Addendum: In the comments, you asked about counting the number of circular seating arrangements of five females and five males if no two males are adjacent.

In a circular arrangement, the convention is that two arrangements are considered equivalent if one can be obtained from the other by rotation. One way to handle equivalence under rotation is to use a particular person (or object) as a reference point.

In a circular arrangement, the only way for five females and five males to be arranged so that no two of them are adjacent is if the females and males alternate seats.

Suppose Amanda is one of the females. Seat her somewhere. It does not matter where since we only care about the relative order of the people. Seating her determines which seats will be occupied by females and which seats will be occupied by males. Once Amanda has been seated, there are $4!$ ways to arrange the remaining females in the four other seats which are now designated for females as you proceed clockwise around the table from Amanda. The five males can be seated in the five remaining seats in $5!$ ways as we proceed clockwise around the table from Amanda. Thus, there are $4!5!$ ways to arrange five females and five males around a circular table if no two males are adjacent.

N. F. Taussig
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  • What if we were seating 5 males and 5 females at a circular table? In this case, I feel like my arguement above would go through exactly? However, the correct answer in this case is $4!5!$... Perhaps I'm supposed to take into account the symmetry of the table...? –  Apr 20 '22 at 22:21
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    I have added an addendum to discuss what happens in a circular arrangement. I discussed circular permutations here. – N. F. Taussig Apr 20 '22 at 23:17
  • Okay, thanks!! In my head, I was thinking that this circular table had 10 "bucket seats". Thanks for all of this, you are the bomb!! –  Apr 20 '22 at 23:50